What is the initial velocity of a ball falling in Earth's gravity?

[Orson]

Homework Statement

The distance S(t) from the ground of a ball falling in the Earth's gravity after being thrown upward with certain initial velocity of 2 ft/sec at the height of 5 feet can be written as:

S(t) = at^2 + 2t + c

Verify whether S(t) given correctly indicate the initial velocity of 2 ft/sec

Homework Equations

S(t) = at^2 + 2t + c

The Attempt at a Solution

I took the derivative of S(t) = at^2 + 2t + c and got 2at + 2. Plugging in zero for t we get 2. (I am assuming initial velocity is at t=0?

 

[MarkFL]

Yes, the initial velocity is generally taken to be:

##v_0=v(0)=s'(0)##

 

[NFuller]

But what happens when you take the second derivative?

 

[Orson]

Yes, the initial velocity is generally taken to be:

##v_0=v(0)=s'(0)##

Assuming only gravity is acting on the falling ball, find the value of a. I think it would be -9.81?

This is actually a calculus 1 class and i have very little physics knowledge, but i filed it here as it involves physics/

But what happens when you take the second derivative?

2a.

 

[NFuller]

2a.

Right, so is the equation written correctly?

 

[Orson]

Assuming only gravity is acting on the falling ball, find the value of a. I think it would be -9.81?

This is actually a calculus 1 class and i have very little physics knowledge, but i filed it here as it involves physics/

2a.

9.81x3.2

 

[Orson]

Right, so is the equation written correctly?

not sure what you mean.

 

[NFuller]

not sure what you mean.

The definition of acceleration is
$$a=\frac{d^2}{dt^2}S(t)$$
When you took the second derivative of ##S(t)## you got
$$\frac{d^2}{dt^2}S(t)=2a$$
So is the equation for ##S(t)## valid?

 

[Orson]

The definition of acceleration is
$$a=\frac{d^2}{dt^2}S(t)$$
When you took the second derivative of ##S(t)## you got
$$\frac{d^2}{dt^2}S(t)=2a$$
So is the equation for ##S(t)## valid?

seems to work for me. I've gone backwards and forwards.

 

[NFuller]

seems to work for me. I've gone backwards and forwards.

The point is that there is a contradiction ##a\ne2a##.

 

[Orson]

The point is that there is a contradiction ##a\ne2a##.

true. another question reads assuming gravity is the only force acting on the ball, what is the value of a? -9.81*3.28 (converting meters into feet) .

then of course they want to know the maximum height, the time it achieves maximum height, and the time it hits the ground. (quadratic time)

but what do i do about this contradiction?

 

[NFuller]

but what do i do about this contradiction?

Well, the question says to verify whether ##S(t)## us given correctly. Clearly its not because the second derivative does not give the acceleration. If you need to write it in the correct form, then inserting a factor of ##1/2## in front of ##a## in the equation for ##S(t)## will fix the contradiction.

 

[Orson]

Well, the question says to verify whether ##S(t)## us given correctly. Clearly its not because the second derivative does not give the acceleration. If you need to write it in the correct form, then inserting a factor of ##1/2## in front of ##a## in the equation for ##S(t)## will fix the contradiction.

was this a trick question?

 

[NFuller]

was this a trick question?

I think the question is worded badly but from what I can tell its asking if they gave you the correct expression for the position of a falling object. By checking the second derivative we have shown that this is not the proper equation and that is your answer.

 

[NFuller]

The only other thing I can thing of is that the problem is just using bad notation and that ##a## doesn't explicitly mean acceleration but is just an arbitrary constant in the equation. If that is the case then acceleration is related to the ##a## they give you by ##a=2\times\text{acceleration}##.

 

[Orson]

The only other thing I can thing of is that the problem is just using bad notation and that ##a## doesn't explicitly mean acceleration but is just an arbitrary constant in the equation. If that is the case then acceleration is related to the ##a## they give you by ##a=2\times\text{acceleration}##.

I'll find out on Wednesday. I love the prof but there is a language barrier.

 

[MarkFL]

The only other thing I can thing of is that the problem is just using bad notation and that ##a## doesn't explicitly mean acceleration but is just an arbitrary constant in the equation. If that is the case then acceleration is related to the ##a## they give you by ##a=2\times\text{acceleration}##.

That was the way I interpreted the given equation.

 

[Ray Vickson]

The point is that there is a contradiction ##a\ne2a##.

The problem (as reproduced by the OP) does not claim that the acceleration is ##a##; it just writes a formula for ##S(t)## that is quadratic and contains two constants it calls ##c## and ##a##.

 

[Orson]

Later in the problem it asks for the value of a, assuming only gravity is acting on the ball. Given this is a calculus course as opposed to a physics course, should i use -10(3.28), -9.80(3.28), or -9.81(3.28)?

 

[MarkFL]

If gravity is the only force acting on the ball, then we would have:

##s(t)=-\dfrac{g}{2}t^2+v_0t+s_0##

And so:

##a=-\dfrac{g}{2}##

If you are required to give a numeric answer, then I would use:

##g\approx32.2\,\dfrac{\text{ft}}{\text{s}^2}##